A body travels 200 cm in the first two seconds and 220 cm in the next 4 seconds with deceleration. the velocity of the body at the end of the `7^(th)` second is

Options

A. 10 cm/s
B. 5 cm/s
C. 15 cm/s
D. 20 cm/s

Explanation

Let ‘u’ and ‘a’ be the initial velocity and acceleration of the body respectively. For first two second (t = 2 sec), the distance covered is 200 cm.
Now using , ` s = u t + 1/2 at^(2)`
We have , ` 200 = u t + 1/2 at^(2)`
we have , 200 = u (B) + (1/2) a ` (B)^(2)` ........(A)
After four seconds of this journey i.e., after a time
t = 6 sec the distance covered is
200 cm + 220 cm = 420 cm.
Hence 420 = u (6) + (1/2) a`(6)^(2)` ...... (B)
Solving equations (A) and (B),
we get, u = 115 cm/sec and ` a = – 15 cm//sec^(2)`
Now velocity after 7 seconds= u + a t =115 + (–15) 7 = 10 cm/sec
Hence correct answer is (A)